Let the mean and variance of 8 numbers $x, y, 10,12,6,12,4,8$ be $9$ and $9.25$ respectively. If $x > y$, then $3 x-2 y$ is equal to _____________.

$$ \begin{array}{|c|c|c|} \hline x_i & (x_i-\bar{x}) & (x_i-\bar{x})^2 \\ \hline x & x-9 & (x-9)^2 \\ \hline y & y-9 & (y-9)^2 \\ \hline 10 & 1 & 1 \\ \hline 12 & 3 & 9 \\ \hline 6 & -3 & 9 \\ \hline 12 & 3 & 9 \\ \hline 4 & -5 & 25 \\ \hline 8 & -1 & 1 \\ \hline x+y+92 & & (x-9)^2+(y-9)^2+54 \\ \hline \end{array} $$ <br/><br/>Now, mean $(\bar{x})=9$ <br/><br/>$$ \begin{aligned} & \Rightarrow \frac{x+y+52}{8}=9 \\\\ & \Rightarrow x+y=20 \end{aligned} $$ <br/><br/>Also, variance $=9.25$ <br/><br/>$$ \begin{aligned} & \Rightarrow \frac{(x-9)^2+(y-9)^2+54}{8}=9.25 \\\\ & \Rightarrow x^2+y^2+81+81-2 \times 9(x+y)=20 \\\\ & \Rightarrow x^2+y^2-18 \times 20=-142 \\\\ & \Rightarrow x^2+y^2=218 \\\\ & \Rightarrow x^2+(20-x)^2=218 \\\\ & \Rightarrow x^2+400+x^2-40 x=218 \\\\ & \Rightarrow 2 x^2-40 x+182=0 \\\\ & \Rightarrow x=\frac{40 \pm 12}{4} \\\\ & \Rightarrow x=13 \text { or } x=7 \Rightarrow y=7 \text { or } y=13 \\\\ & \text { But } x>y \\\\ & \therefore x=13 \text { and } y=7 \\\\ & \text { So, } 3 x-2 y=39-14=25 \end{aligned} $$ <br/><br/><b>Concept :</b> <br/><br/>(a) Mean $=\frac{\Sigma x_i}{n}$ <br/><br/>(b) Variance $=\frac{\Sigma\left(x_i-\bar{x}\right)^2}{n}$